Question

A vehicle owner interested in her vehicle’s fuel economy calculated the gas mileage in miles per gallon (mpg) each time she filled her tank, for several years. Here are the mpg values for a random sample of 20 of these records. Suppose that the standard deviation is known to be σ = 3.5 mpg.

Required:
Find the 95% confidence interval for the mean of gas mileage for this vehicle.

Answer 1

The 95% confidence interval for the mean of gas mileage for this vehicle is (S - 1.53, S + 1.53), in which S is the mean of the sample of 20.

Explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1 - 0.95)/(2) = 0.025`

Now, we have to find z in the Ztable as such z has a pvalue of
`1 - \alpha`.

That is z with a pvalue of
`1 - 0.025 = 0.975`, so Z = 1.96.

Now, find the margin of error M as such

`M = z(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

`M = 1.96(3.5)/(√(20)) = 1.53`

The lower end of the interval is the sample mean subtracted by M. So it is S - 1.53.

The upper end of the interval is the sample mean added to M. So it is S + 1.53.

The 95% confidence interval for the mean of gas mileage for this vehicle is (S - 1.53, S + 1.53), in which S is the mean of the sample of 20.